In​ 2012, the population of a city was 5.82 million. The exponential growth rate was 2.95% per year.
​a) Find the exponential growth function.
​b) Estimate the population of the city in 2018.
​c) When will the population of the city be ​million?
​d) Find the doubling time.

Answers

Answer 1

Answer:

a) P(t) = 6.57e ^{0.0287t}P(t)=6.57e

0.0287t

b) P(6) = 7,805\ millionP(6)=7,805 million

c) t = 14.64\ yearst=14.64 years

d) t = 24.15\ yearst=24.15 years

Step-by-step explanation:

a) The function of exponential growth for a population has the following formula:

P(t) = p_0e ^{rt}P(t)=p

0

e

rt

In this equation:

p_0p

0

is the initial population

r is the growth rate

t is the time in years

In this problem we know that

r = 2.87\% = 0.0287r=2.87%=0.0287

p_0= 6.57p

0

=6.57 million in the year 2012.

So the equation is:

P(t) = 6.57e ^{0.0287t}P(t)=6.57e

0.0287t

Where t = 0t=0 represents the year 2012

b) If t = 0t=0 in 2012, then in 2018 t = 6t=6

The population in 2018 is:

\begin{gathered}P(t = 6) = 6.57e ^{0.0287(6)}\\\\P(6) = 7,805\ million\end{gathered}

P(t=6)=6.57e

0.0287(6)

P(6)=7,805 million

c) To know when the population is equal to 10 million we must equal P(t) to 10 and solve for t.

\begin{gathered}P(t) = 10 = 6.57e ^{0.0287t}\\\\\frac{10}{6.57} = e ^{0.0287t}\\\\ln(\frac{10}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{10}{6.57})}{0.0287}\end{gathered}

P(t)=10=6.57e

0.0287t

6.57

10

=e

0.0287t

ln(

6.57

10

)=0.0287t

t=

0.0287

ln(

6.57

10

)

t = 14.64\ yearst=14.64 years

d) The function is doubled when P(t) = 2p_0P(t)=2p

0

P(t) = 2(6.57) = 13.14 = 6.57e ^{0.0287t}P(t)=2(6.57)=13.14=6.57e

0.0287t

We solve for t.

\begin{gathered}\frac{13.14}{6.57} = e ^{0.0287t}\\\\ln(\frac{13.14}{6.57}) = 0.0287t\\\\t = ln{13.14}{6.57})}{0.0287})\

end

{gathered}

6.57

13.14

=e

0.0287t

ln(

6.57

13.14

)=0.0287t

t=

0.0287

ln(

6.57

13.14

)

)

t = 24.15\ yearst=24.15 years


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[tex]{\underline{\pink{\textsf{\textbf{ Answer : }}}}}[/tex]

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[tex]{\underline{\purple{\textsf{\textbf{ Explanation : }}}}}[/tex]

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Order of Operations: BPEMDAS

Geometry

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Step-by-step explanation:

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Answer:

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Answer:

Answer and Explanation:

We have:

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μ

=

3

,

000

hours

Population standard deviation,

σ

=

696

hours

Sample size,

n

=

36

1) The standard deviation of the sampling distribution:

σ

¯

x

=

σ

n

=

696

36

=

116

2) As per the central limit theorem, the expected value of the sampling distribution is equal to the population mean.

Therefore:

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μ

¯

x

=

μ

=

3

,

000

The standard deviation of the sampling distribution,

σ

¯

x

=

116

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¯

x

is approximately normal. As the sample size is more than

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.

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and

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P

(

2670.56

<

x

<

2809.76

)

=

P

(

2670.56

3000

116

<

z

<

2809.76

3000

116

)

=

P

(

2.84

<

z

<

1.64

)

=

P

(

z

<

1.64

)

P

(

z

<

2.84

)

=

0.0482

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4) The probability that the average life in the sample will be greater than

3219.24

hours:

P

(

x

>

3219.24

)

=

P

(

z

>

3219.24

3000

116

)

=

P

(

z

>

1.89

)

=

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P

(

x

<

3180.96

)

=

P

(

z

<

3180.96

3000

116

)

=

P

(

z

<

1.56

)

=

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Given that;

The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours.

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Now,

Since, The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours.

And, A simple random sample of 36 bulbs is taken.

Hence, The expected value, standard deviation, and shape of the sampling distribution of is,

⇒ Standard error = 696 / √ 36

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https://brainly.com/question/475676

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[tex] \frac{1}{8} = \frac{1 \times 9}{8 \times 9} = \frac{9}{72} \\ [/tex]

_________________________________

[tex] \frac{8}{72} < \frac{9}{72} < \frac{36}{72} \\ [/tex]

[tex] \frac{1}{9} < \frac{1}{8} < \frac{1}{2} \\ [/tex]

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Thus the order from greatest to least is :

[tex] \frac{1}{2} \: , \: \frac{1}{8} \: , \: \frac{1}{9} \\ [/tex]

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Answer:

X = 16

Step-by-step explanation:

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32 = x + 16

-x = -16

so x = 16

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Mark it as Brainliest .

Answer:

x=16

Step-by-step explanation:

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Answers

Answer:

The equation is C(t) = 1.50t + 4.00 and the data table is below.

Step-by-step explanation:

We are given that a cafe charges $1.50 for every minute that someone uses the internet. The cafe also charges an initial $4 fee to use the internet.

We need to find a function, in terms of t, that will find these values.

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Therefore, we can set up our function.

C is the total cost and t is the time, in minutes, that internet is used at the cafe. Our equation can be set up as [tex]C(t) = 1.50t + 4.00[/tex].

Now, we can create a table of values for t = 0 to t = 5 to determine what C(t) is at each of these prime intervals.

[tex]\begin{array}{|c|c|} \cline{1-2} \textbf{C} & \textbf{t} \\ \cline{1-2} a & 0 \\ \cline{1-2} b & 1 \\ \cline{1-2} c & 2 \\ \cline{1-2} d & 3 \\ \cline{1-2} e & 4 \\ \cline{1-2} f & 5 \\ \cline{1-2} \end{array}[/tex]

This is what our table looks like, but we need to fill the C column. Therefore, we can use substitution of our t-values into the equation and solve for C.

[tex]\bullet \ \text{For t = 0,}[/tex]

[tex]C(0) = 1.50(0) + 4\\\\C(0) = 0 + 4\\\\C(0) = 4.00[/tex]

[tex]\bullet \ \text{For t = 1,}[/tex]

[tex]C(1) = 1.50(1) +4\\\\C(1) = 1.50 + 4\\\\C(1) = 5.50[/tex]

[tex]\bullet \ \text{For t = 2,}[/tex]

[tex]C(2) = 1.50(2) + 4\\\\C(2) = 3.00 + 4\\\\C(2) = 7.00[/tex]

[tex]\bullet \ \text{For t = 3,}[/tex]

[tex]C(3) = 1.50(3) + 4\\\\C(3) = 4.50 + 4\\\\C(3) = 8.50[/tex]

[tex]\bullet \ \text{For t = 4,}[/tex]

[tex]C(4) = 1.50(4) + 4\\\\C(4) = 6.00 + 4\\\\C(4) = 10.00[/tex]

[tex]\bullet \ \text{For t = 5,}[/tex]

[tex]C(5) = 1.50(5) + 4\\\\C(5) = 7.50 + 4\\\\C(5) = 11.50[/tex]

Now, we can insert these values into our table.

[tex]\begin{array}{|c|c|} \cline{1-2} \textbf{C} & \textbf{t} \\ \cline{1-2} 4.00 & 0 \\ \cline{1-2} 5.50 & 1 \\ \cline{1-2} 7.00 & 2 \\ \cline{1-2} 8.50 & 3 \\ \cline{1-2} 10.00 & 4 \\ \cline{1-2} 11.50 & 5 \\ \cline{1-2} \end{array}[/tex]

Therefore, our equation is C(t) = 1.50t + 4.00 and our data table is above.

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Answers

Answer:

400-40x x=6

Step-by-step explanation:

400 in total. minus 40 each week (x)

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